Question: As a particle moves along the number line, its position at time $t$ is $s(t)$, its velocity is $v(t)$, and its acceleration is $a(t)= 1$. If $v(3) = -3$ and $s(2) = -10$, what is $s(4)$ ? $s(4)=~$
Answer: The antiderivative of $~a(t)~$ is $~v(t)=t+C\,$. We know that $~v(3) =-3\,$, so $~C=-6\,$. Therefore, $~v(t) = t-6\,$. The antiderivative of $~v(t)~$ is $~s(t)=\dfrac12t^2-6t+K\,$. We know that $~s(2) = -10\,$, so $~K=0\,$. Therefore $~s(t)=\dfrac12t^2-6t\,$. Then $~s(4)=8-24=-16\,$.